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    "<div style=\"color:#777777;background-color:#ffffff;font-size:12px;text-align:right;\">\n",
    "\tprepared by Abuzer Yakaryilmaz (QuSoft@Riga) | November 02, 2018\n",
    "</div>\n",
    "<table><tr><td><i> I have some macros here. If there is a problem with displaying mathematical formulas, please run me to load these macros.</i></td></td></table>\n",
    "$ \\newcommand{\\bra}[1]{\\langle #1|} $\n",
    "$ \\newcommand{\\ket}[1]{|#1\\rangle} $\n",
    "$ \\newcommand{\\braket}[2]{\\langle #1|#2\\rangle} $\n",
    "$ \\newcommand{\\inner}[2]{\\langle #1,#2\\rangle} $\n",
    "$ \\newcommand{\\biginner}[2]{\\left\\langle #1,#2\\right\\rangle} $\n",
    "$ \\newcommand{\\mymatrix}[2]{\\left( \\begin{array}{#1} #2\\end{array} \\right)} $\n",
    "$ \\newcommand{\\myvector}[1]{\\mymatrix{c}{#1}} $\n",
    "$ \\newcommand{\\myrvector}[1]{\\mymatrix{r}{#1}} $\n",
    "$ \\newcommand{\\mypar}[1]{\\left( #1 \\right)} $\n",
    "$ \\newcommand{\\mybigpar}[1]{ \\Big( #1 \\Big)} $\n",
    "$ \\newcommand{\\sqrttwo}{\\frac{1}{\\sqrt{2}}} $\n",
    "$ \\newcommand{\\dsqrttwo}{\\dfrac{1}{\\sqrt{2}}} $\n",
    "$ \\newcommand{\\onehalf}{\\frac{1}{2}} $\n",
    "$ \\newcommand{\\donehalf}{\\dfrac{1}{2}} $\n",
    "$ \\newcommand{\\hadamard}{ \\mymatrix{rr}{ \\sqrttwo & \\sqrttwo \\\\ \\sqrttwo & -\\sqrttwo }} $\n",
    "$ \\newcommand{\\vzero}{\\myvector{1\\\\0}} $\n",
    "$ \\newcommand{\\vone}{\\myvector{0\\\\1}} $\n",
    "$ \\newcommand{\\vhadamardzero}{\\myvector{ \\sqrttwo \\\\  \\sqrttwo } } $\n",
    "$ \\newcommand{\\vhadamardone}{ \\myrvector{ \\sqrttwo \\\\ -\\sqrttwo } } $\n",
    "$ \\newcommand{\\myarray}[2]{ \\begin{array}{#1}#2\\end{array}} $\n",
    "$ \\newcommand{\\X}{ \\mymatrix{cc}{0 & 1 \\\\ 1 & 0}  } $\n",
    "$ \\newcommand{\\Z}{ \\mymatrix{rr}{1 & 0 \\\\ 0 & -1}  } $\n",
    "$ \\newcommand{\\Htwo}{ \\mymatrix{rrrr}{ \\frac{1}{2} & \\frac{1}{2} & \\frac{1}{2} & \\frac{1}{2} \\\\ \\frac{1}{2} & -\\frac{1}{2} & \\frac{1}{2} & -\\frac{1}{2} \\\\ \\frac{1}{2} & \\frac{1}{2} & -\\frac{1}{2} & -\\frac{1}{2} \\\\ \\frac{1}{2} & -\\frac{1}{2} & -\\frac{1}{2} & \\frac{1}{2} } } $\n",
    "$ \\newcommand{\\CNOT}{ \\mymatrix{cccc}{1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0} } $\n",
    "$ \\newcommand{\\norm}[1]{ \\left\\lVert #1 \\right\\rVert } $"
   ]
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h2> <font color=\"blue\"> Solutions for </font> Basics of Python: Lists </h2>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id=\"task1\"></a>\n",
    "<h3> Task 1 </h3>\n",
    "\n",
    "Fibonacci sequence starts with $ 1 $ and $ 1 $. Then, each next element is equal to summation of the previous two elements:\n",
    "$$\n",
    "    1, 1, 2 , 3 , 5, 8, 13, 21, 34, 55 \\ldots \n",
    "$$\n",
    "\n",
    "Find the first 30 elements of the Fibonacci sequence, store them in a list, and then print the list. \n",
    "\n",
    "You can verify the first 10 elements of your result with the above list."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h3>Solution</h3>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# the first and second elements are 1 and 1\n",
    "F = [1,1]\n",
    "\n",
    "for i in range(2,30):\n",
    "    F.append(F[i-1] + F[i-2])\n",
    "\n",
    "# print the final list\n",
    "print(F)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<a id=\"task2\"></a>\n",
    "<h3> Task 2 </h3>\n",
    "\n",
    "Define a list $ N $ with 11 elements such that $ N[i] $ is another list with four elements as $ [i,i^2,i^3,i^2+i^3] $.\n",
    "\n",
    "The index $ i $ should be form $ 0 $ and $ 10 $."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<h3>Solution</h3>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# define an empty list\n",
    "N = []\n",
    "\n",
    "for i in range(11):\n",
    "    N.append([ i , i*i , i*i*i , i*i + i*i*i ]) # a list having four elements is added to the list N\n",
    "    # Alternatively:\n",
    "    #N.append([i , i**2 , i**3 , i**2 + i**3]) # ** is the exponent operator\n",
    "    #N = N + [ [i , i*i , i*i*i , i*i + i*i*i] ] # Why using double brakets?\n",
    "    #N = N + [ [i , i**2 , i**3 , i**2 + i**3] ] # Why using double brakets?\n",
    "    # In the last two alternative solutions, you may try with a single braket, \n",
    "    #     and then see why double brakets are needed for the exact solution.\n",
    "\n",
    "# print the final list\n",
    "print(N)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# let's print the list element by element\n",
    "for i in range(len(N)):\n",
    "    print(N[i])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# let's print the list element by element by using an alternative method\n",
    "for el in N: # el will iteratively takes the values of elements in N \n",
    "    print(el)"
   ]
  }
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